Integrand size = 9, antiderivative size = 86 \[ \int x^4 \operatorname {PolyLog}(2,a x) \, dx=-\frac {x}{25 a^4}-\frac {x^2}{50 a^3}-\frac {x^3}{75 a^2}-\frac {x^4}{100 a}-\frac {x^5}{125}-\frac {\log (1-a x)}{25 a^5}+\frac {1}{25} x^5 \log (1-a x)+\frac {1}{5} x^5 \operatorname {PolyLog}(2,a x) \] Output:
-1/25*x/a^4-1/50*x^2/a^3-1/75*x^3/a^2-1/100*x^4/a-1/125*x^5-1/25*ln(-a*x+1 )/a^5+1/25*x^5*ln(-a*x+1)+1/5*x^5*polylog(2,a*x)
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.85 \[ \int x^4 \operatorname {PolyLog}(2,a x) \, dx=\frac {-a x \left (60+30 a x+20 a^2 x^2+15 a^3 x^3+12 a^4 x^4\right )+60 \left (-1+a^5 x^5\right ) \log (1-a x)+300 a^5 x^5 \operatorname {PolyLog}(2,a x)}{1500 a^5} \] Input:
Integrate[x^4*PolyLog[2, a*x],x]
Output:
(-(a*x*(60 + 30*a*x + 20*a^2*x^2 + 15*a^3*x^3 + 12*a^4*x^4)) + 60*(-1 + a^ 5*x^5)*Log[1 - a*x] + 300*a^5*x^5*PolyLog[2, a*x])/(1500*a^5)
Time = 0.28 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {7145, 25, 2842, 49, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^4 \operatorname {PolyLog}(2,a x) \, dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {1}{5} x^5 \operatorname {PolyLog}(2,a x)-\frac {1}{5} \int -x^4 \log (1-a x)dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{5} \int x^4 \log (1-a x)dx+\frac {1}{5} x^5 \operatorname {PolyLog}(2,a x)\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{5} a \int \frac {x^5}{1-a x}dx+\frac {1}{5} x^5 \log (1-a x)\right )+\frac {1}{5} x^5 \operatorname {PolyLog}(2,a x)\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{5} a \int \left (-\frac {x^4}{a}-\frac {x^3}{a^2}-\frac {x^2}{a^3}-\frac {x}{a^4}-\frac {1}{a^5 (a x-1)}-\frac {1}{a^5}\right )dx+\frac {1}{5} x^5 \log (1-a x)\right )+\frac {1}{5} x^5 \operatorname {PolyLog}(2,a x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{5} a \left (-\frac {\log (1-a x)}{a^6}-\frac {x}{a^5}-\frac {x^2}{2 a^4}-\frac {x^3}{3 a^3}-\frac {x^4}{4 a^2}-\frac {x^5}{5 a}\right )+\frac {1}{5} x^5 \log (1-a x)\right )+\frac {1}{5} x^5 \operatorname {PolyLog}(2,a x)\) |
Input:
Int[x^4*PolyLog[2, a*x],x]
Output:
((x^5*Log[1 - a*x])/5 + (a*(-(x/a^5) - x^2/(2*a^4) - x^3/(3*a^3) - x^4/(4* a^2) - x^5/(5*a) - Log[1 - a*x]/a^6))/5)/5 + (x^5*PolyLog[2, a*x])/5
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_ ))^(q_.), x_Symbol] :> Simp[(f + g*x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/( g*(q + 1))), x] - Simp[b*e*(n/(g*(q + 1))) Int[(f + g*x)^(q + 1)/(d + e*x ), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]
Int[((d_.)*(x_))^(m_.)*PolyLog[n_, (a_.)*((b_.)*(x_)^(p_.))^(q_.)], x_Symbo l] :> Simp[(d*x)^(m + 1)*(PolyLog[n, a*(b*x^p)^q]/(d*(m + 1))), x] - Simp[p *(q/(m + 1)) Int[(d*x)^m*PolyLog[n - 1, a*(b*x^p)^q], x], x] /; FreeQ[{a, b, d, m, p, q}, x] && NeQ[m, -1] && GtQ[n, 0]
Time = 0.65 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.84
method | result | size |
meijerg | \(\frac {-\frac {a x \left (12 a^{4} x^{4}+15 x^{3} a^{3}+20 a^{2} x^{2}+30 x a +60\right )}{1500}-\frac {\left (-6 x^{5} a^{5}+6\right ) \ln \left (-x a +1\right )}{150}+\frac {x^{5} a^{5} \operatorname {polylog}\left (2, x a \right )}{5}}{a^{5}}\) | \(72\) |
parallelrisch | \(\frac {300 x^{5} a^{5} \operatorname {polylog}\left (2, x a \right )+60 \ln \left (-x a +1\right ) x^{5} a^{5}-12 x^{5} a^{5}-15 a^{4} x^{4}-60-20 x^{3} a^{3}-30 a^{2} x^{2}-60 x a -60 \ln \left (-x a +1\right )}{1500 a^{5}}\) | \(81\) |
parts | \(\frac {x^{5} \operatorname {polylog}\left (2, x a \right )}{5}-\frac {\frac {\left (-x a +1\right )^{5} \ln \left (-x a +1\right )}{5}-\frac {\left (-x a +1\right )^{5}}{25}-\ln \left (-x a +1\right ) \left (-x a +1\right )^{4}+\frac {\left (-x a +1\right )^{4}}{4}+2 \ln \left (-x a +1\right ) \left (-x a +1\right )^{3}-\frac {2 \left (-x a +1\right )^{3}}{3}-2 \ln \left (-x a +1\right ) \left (-x a +1\right )^{2}+\left (-x a +1\right )^{2}+\ln \left (-x a +1\right ) \left (-x a +1\right )+x a -1}{5 a^{5}}\) | \(142\) |
derivativedivides | \(\frac {\frac {x^{5} a^{5} \operatorname {polylog}\left (2, x a \right )}{5}-\frac {\left (-x a +1\right )^{5} \ln \left (-x a +1\right )}{25}+\frac {\left (-x a +1\right )^{5}}{125}+\frac {\ln \left (-x a +1\right ) \left (-x a +1\right )^{4}}{5}-\frac {\left (-x a +1\right )^{4}}{20}-\frac {2 \ln \left (-x a +1\right ) \left (-x a +1\right )^{3}}{5}+\frac {2 \left (-x a +1\right )^{3}}{15}+\frac {2 \ln \left (-x a +1\right ) \left (-x a +1\right )^{2}}{5}-\frac {\left (-x a +1\right )^{2}}{5}-\frac {\ln \left (-x a +1\right ) \left (-x a +1\right )}{5}+\frac {1}{5}-\frac {x a}{5}}{a^{5}}\) | \(147\) |
default | \(\frac {\frac {x^{5} a^{5} \operatorname {polylog}\left (2, x a \right )}{5}-\frac {\left (-x a +1\right )^{5} \ln \left (-x a +1\right )}{25}+\frac {\left (-x a +1\right )^{5}}{125}+\frac {\ln \left (-x a +1\right ) \left (-x a +1\right )^{4}}{5}-\frac {\left (-x a +1\right )^{4}}{20}-\frac {2 \ln \left (-x a +1\right ) \left (-x a +1\right )^{3}}{5}+\frac {2 \left (-x a +1\right )^{3}}{15}+\frac {2 \ln \left (-x a +1\right ) \left (-x a +1\right )^{2}}{5}-\frac {\left (-x a +1\right )^{2}}{5}-\frac {\ln \left (-x a +1\right ) \left (-x a +1\right )}{5}+\frac {1}{5}-\frac {x a}{5}}{a^{5}}\) | \(147\) |
Input:
int(x^4*polylog(2,x*a),x,method=_RETURNVERBOSE)
Output:
1/a^5*(-1/1500*a*x*(12*a^4*x^4+15*a^3*x^3+20*a^2*x^2+30*a*x+60)-1/150*(-6* a^5*x^5+6)*ln(-a*x+1)+1/5*x^5*a^5*polylog(2,x*a))
Time = 0.12 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.84 \[ \int x^4 \operatorname {PolyLog}(2,a x) \, dx=\frac {300 \, a^{5} x^{5} {\rm Li}_2\left (a x\right ) - 12 \, a^{5} x^{5} - 15 \, a^{4} x^{4} - 20 \, a^{3} x^{3} - 30 \, a^{2} x^{2} - 60 \, a x + 60 \, {\left (a^{5} x^{5} - 1\right )} \log \left (-a x + 1\right )}{1500 \, a^{5}} \] Input:
integrate(x^4*polylog(2,a*x),x, algorithm="fricas")
Output:
1/1500*(300*a^5*x^5*dilog(a*x) - 12*a^5*x^5 - 15*a^4*x^4 - 20*a^3*x^3 - 30 *a^2*x^2 - 60*a*x + 60*(a^5*x^5 - 1)*log(-a*x + 1))/a^5
Time = 2.77 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.77 \[ \int x^4 \operatorname {PolyLog}(2,a x) \, dx=\begin {cases} - \frac {x^{5} \operatorname {Li}_{1}\left (a x\right )}{25} + \frac {x^{5} \operatorname {Li}_{2}\left (a x\right )}{5} - \frac {x^{5}}{125} - \frac {x^{4}}{100 a} - \frac {x^{3}}{75 a^{2}} - \frac {x^{2}}{50 a^{3}} - \frac {x}{25 a^{4}} + \frac {\operatorname {Li}_{1}\left (a x\right )}{25 a^{5}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:
integrate(x**4*polylog(2,a*x),x)
Output:
Piecewise((-x**5*polylog(1, a*x)/25 + x**5*polylog(2, a*x)/5 - x**5/125 - x**4/(100*a) - x**3/(75*a**2) - x**2/(50*a**3) - x/(25*a**4) + polylog(1, a*x)/(25*a**5), Ne(a, 0)), (0, True))
Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.84 \[ \int x^4 \operatorname {PolyLog}(2,a x) \, dx=\frac {300 \, a^{5} x^{5} {\rm Li}_2\left (a x\right ) - 12 \, a^{5} x^{5} - 15 \, a^{4} x^{4} - 20 \, a^{3} x^{3} - 30 \, a^{2} x^{2} - 60 \, a x + 60 \, {\left (a^{5} x^{5} - 1\right )} \log \left (-a x + 1\right )}{1500 \, a^{5}} \] Input:
integrate(x^4*polylog(2,a*x),x, algorithm="maxima")
Output:
1/1500*(300*a^5*x^5*dilog(a*x) - 12*a^5*x^5 - 15*a^4*x^4 - 20*a^3*x^3 - 30 *a^2*x^2 - 60*a*x + 60*(a^5*x^5 - 1)*log(-a*x + 1))/a^5
\[ \int x^4 \operatorname {PolyLog}(2,a x) \, dx=\int { x^{4} {\rm Li}_2\left (a x\right ) \,d x } \] Input:
integrate(x^4*polylog(2,a*x),x, algorithm="giac")
Output:
integrate(x^4*dilog(a*x), x)
Time = 0.23 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.80 \[ \int x^4 \operatorname {PolyLog}(2,a x) \, dx=\frac {x^5\,\ln \left (1-a\,x\right )}{25}-\frac {\ln \left (a\,x-1\right )}{25\,a^5}-\frac {x}{25\,a^4}-\frac {x^5}{125}+\frac {x^5\,\mathrm {polylog}\left (2,a\,x\right )}{5}-\frac {x^4}{100\,a}-\frac {x^3}{75\,a^2}-\frac {x^2}{50\,a^3} \] Input:
int(x^4*polylog(2, a*x),x)
Output:
(x^5*log(1 - a*x))/25 - log(a*x - 1)/(25*a^5) - x/(25*a^4) - x^5/125 + (x^ 5*polylog(2, a*x))/5 - x^4/(100*a) - x^3/(75*a^2) - x^2/(50*a^3)
\[ \int x^4 \operatorname {PolyLog}(2,a x) \, dx=\int \mathit {polylog}\left (2, a x \right ) x^{4}d x \] Input:
int(x^4*polylog(2,a*x),x)
Output:
int(polylog(2,a*x)*x**4,x)